Desenăm triunghiul oarecare ABC.
Ducem DE || BC, cu D mai aproape de A decât de B, pentru că
din enunț rezultă AD < DB.
[tex]\it DE||BC\ \stackrel{T.f.a.}{\Longrightarrow} \Delta ADE\sim\Delta ABC\Rightarrow\dfrac{AD}{AB}=\dfrac{DE}{BC}\ \stackrel{derivare}{\Longrightarrow}\\ \\ \\ \Rightarrow\dfrac{AD}{AB-AD}=\dfrac{DE}{BC-DE}\Rightarrow\dfrac{AD}{DB}=\dfrac{DE}{25-DE}\Rightarrow\dfrac{2}{3}=\dfrac{DE}{25-DE}\Rightarrow\\ \\ \\ \Rightarrow\dfrac{3}{2}=\dfrac{25-DE}{DE}\Rightarrow\dfrac{3}{2}=\dfrac{25}{DE}-\dfrac{DE}{DE}\Rightarrow \dfrac{3}{2}=\dfrac{25}{DE}-1\Rightarrow\dfrac{3}{2}+1=\dfrac{25}{DE}\Rightarrow[/tex]
[tex]\it\Rightarrow\dfrac{^{5)}5}{\ 2}=\dfrac{25}{DE}\Rightarrow\dfrac{25}{10}=\dfrac{25}{DE}\Rightarrow DE=10\ cm[/tex]
